Institute for Theoretical Physics
Heidelberg University
Winter term 2012/13
String Theory and String Phenomenology
Prof. Dr. Arthur Hebecker
Sebastian Kraus
Problem Sheet 12
(Due: January 23, 2013)
Problem 12.1:
Consider a torus T 2 (which is topologically the same as S 1 × S 1 ), parametrised by
for m, n ∈ Z.
(σ 1 , σ 2 ) ' (σ 1 , σ 2 ) + 2π(m, n)
(1)
(a) Convince yourself that, up to an overall rescaling, the most general metric is given by
ds2 = |dσ 1 + τ dσ 2 |2 ,
for τ ∈ C.
(2)
(b) Argue that Weyl transformations and diffeomorphisms can be used to define new coordinates ξ i
in terms of which the metric takes the standard form ds2 = |dξ 1 + idξ 2 |2 .
√
R
1
2 ξ h R = 2 − 2g = 0.
Hint: Use that χ = 4π
d
2
T
Argue that the new periodicities are given by
ξ 1 ' ξ 1 + 2π(m + nv 1 ),
ξ 2 ' ξ 2 + 2π(nv 2 ),
for some vector (v 1 , v 2 ).
(c) Now define the complex variable w = σ 1 + τ σ 2 such that in the first picture ds2 = dwdw.
Convince yourself that the identification (1) implies the identification
w ' w + 2π(m + nτ ),
(3)
i.e. a T 2 is a lattice in C.
Remark: The upshot is the following: We can parametrize the torus in two equivalent ways: In
the first picture (1), the coordinates σ 1 , σ 2 have standard periodicity 2π, but there appears a
modulus τ - the so-called Teichmüller parameter - in the metric (2). Alternatively, we can choose
coordinates w, w such that the metric has the standard flat form ds2 = dwdw, but now in general
the periodicity (3) contains a modulus τ that cannot be removed by conformal transformations.
(d) Consider now the transformations
T :τ →τ +1
1
S:τ →− ,
τ
U :τ →
τ
.
τ +1
Determine their action on the parameters (m, n) in (3) and use this to argue that T , S and U
leave the torus invariant. Argue that T , S and U generate the group SL(2, Z), whose action on
τ is
aτ + b
τ→
,
ad − bc = 1,
a, b, c, d ∈ Z.
cτ + d
Argue that the full modular group of T 2 is P SL(2, Z) = SL(2, Z)/Z2 .
Note: In fact, S and T suffice to generate SL(2, Z)/Z, but this is non-trivial to show.
(e) Use P SL(2, Z) invariance to show that τ can always be brought to the fundamental domain
1
1
F0 : − ≤ <(τ ) ≤ ,
2
2
|τ | ≥ 1.
Represent F0 graphically and discuss the various identifications of its boundaries.
1
Problem 12.2:
In this exercise we want to compute the one-loop vacuum amplitude in the oriented closed string
theory. The corresponding worldsheet is a torus and there are no vertex operator insertions. The
moduli space of the torus was subject of the discussion in problem 12.1. The upshot of this discussion
was that (unlike in the case of the tree-level amplitude on S 2 discussed in the lecture, which has no
metric moduli) conformally inequivalent tori are characterized by a complex parameter τ which
R takes
1
values in the fundamental domain F0 . Thus, the Amplitude involves an integration over 2 F0 d2 τ
where the factor 12 appears due to an extra Z2 symmetry which sends the complex coordinate w on
the torus to its negative value, w → −w.
The conformal Killing group of the torus is U(1)×U(1), corresponding to constant shifts along the
coordinate axes σ 1 and σ 2 .
(a) Show that the volume of the conformal Killing group of the torus is (2π)2 =(τ ) ≡ (2π)2 τ2 .
To avoid overcounting we have to divide by this volume factor. It is clear that, in analogy to the
tree-level amplitude computed in the lecture, the one-loop amplitude will factorize into a product of
independent X- and (bc)-parts which are, in the end, integrated over the moduli space of the torus.
More precisely, the amplitude takes the form
Z
dτ dτ
ZT 2 :=
h1iT 2 (τ ),X · hghost insertionsiT 2 (τ ),bc
F0 4τ2
R
where h1iT 2 (τ ),X denotes the path integral DXe−SX on a worldsheet given by a torus with modulus
τ , while hghost insertionsiT 2 (τ ),bc denotes the corresponding path integral in the (bc)-ghost sector on
the same torus with some ghost field insertions. The latter can be computed to be (cf. Polchinski, vol.
1, page 212ff)
hghost insertionsiT 2 (τ ),bc = hb(0)b̃(0)c̃(0)c(0)iT 2 (τ )
In this expression the two b insertions come from the one (complex) metric modulus. In order to
understand the appearance of the c insertions, think of first computing a one-loop amplitude with an
arbitrary number of vertex operator insertions. Then, as you know, the existence of the two conformal
Killing vectors discussed above allows to fix the position of one vertex operator. This gauge fixing
introduces the c fields. However, since the volume of the conformal Killing group is finite, one can
alternatively reintroduce the integration over the position of the vertex operator insertion (which we
fixed previously) and add a compensating volume factor in the denominator. All vertex operators are
now on equal footing. To obtain the vacuum amplitude one then just takes this expression with the
c fields and omits the vertex operator insertions. The fact that all ghosts are inserted at the origin
comes about because the ghost path integral turns out to be independent of the position of the ghost
fields.
We will now only be concerned with the computation of
Z(τ ) := h1iT 2 (τ ),X .
R
One can think of the path integral DXe−SX on the torus with modulus τ as being the same as the
sum over all states of the X µ -field theory on a circle which are evolved for an Euclidean time 2πτ2 ,
translated in σ 1 direction by 2πτ1 and then identified with themselves.
(b) Show that, in the canonical formalism, this translates into
Z(τ ) = Tr e2πiτ1 P −2πτ2 H
d
= (qq)− 24 Tr q L0 q L0 , q ≡ e2πiτ ,
where, as usual, P = L0 − L0 generates translations along σ 1 , while H = L0 + L0 − c+c
24 generates
time translations.
2
(c) Express
L0 and L0 in terms
momentum and number operators. Furthermore, recall that
R of
P
−d
d
−−−−−−−−−→ Vd (2π)
d k where Vd ≡ ‘spacetime volume’ to arrive at
k −
continuum limit
Z(τ ) = Vd (qq)
d
− 24
dd k −πτ2 α0 k2 0 N N
e
Tr q q ,
(2π)d
Z
where Tr0 denotes the trace over the oscillator part of the string states only.
(d) Show that
Tr0 q N q N =
∞
Y
(1 − q n )−d (1 − q n )−d
n=1
(e) Perform a Wick rotation k 0 → ik 0 to render the integral
order to obtain the final result
R
dd k finite. Compute the integral in
Z(τ ) = iVd (ZX (τ ))d ,
− 1
ZX (τ ) = 4π 2 α0 τ2 2 · |η(τ )|−2 with
∞
1 Y
η(τ ) = q 24
(1 − q n ) the Dedekind η-function.
n=1
The calculation of the ghost contribution to the one-loop amplitude can be found e.g. in Polchinski,
vol. 1, page 212ff. As expected, this contribution cancels the oscillator contributions from the two
non-transverse polarizations, i.e. it yields a factor |η(τ )|4 . Thus, setting d = 26, the one-loop vacuum
amplitude reads
Z
−13
dτ dτ
ZT 2 = iV26
4π 2 α0 τ2
· |η(τ )|−48 .
4τ
2
F0
3